Jan 6 Parametric Equations and Derivatives

Parametric Form of the Derivative

First Derivative

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Let's say you have the parametric equation $$\begin{cases}x = f(t)\\y = g(t)\end{cases}$$
$\dfrac{dy}{dx}$, or the combined derivative (if you were to merge the two equations into one), equals $$\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ or } \dfrac{g'(t)}{f'(t)}$$

Second and Higher Degree Derivatives

The second derivative is the derivative of the first derivative / derivative of x , so it's:

MathJax TeX Test Page $$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \boxed{\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}}$$
The 3rd,4th, 5th, etc. derivatives are almost the same as the second derivative. In the third derivative, you replace $\dfrac{dy}{dx}$ with the second derivative. The idea is the same, you want to find the derivative of that, but you must divide by the derivative of x, because x changes differently than usual (if $x = t^2$, it does not change linearly).

(t/2) /(1/(2√t) = t√t = t3/2

Speed

Sometimes, a problem might ask what the speed of a set of parametric equations is.

Example

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If x(t) = 2t, y(t) = sin(4t), what is the speed at t = 3?
$$x'(t) = 2, y'(t) = 4\cos(4t)$$
We know the x and y speeds, so if you add the two vectors, you get the overall speed.
$$\text{Speed(t)} = \sqrt{(x'(t))^2 + (y'(t))^2}$$

This is the same as finding the tangent to any other curve. You create a line containing that point and the slope. You use point-slope form, so if the point is (a,b), you find what t is and you do y = b + g'(t)/f'(t)(x-a).